A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Some how, the radius of the loop starts expanding at a constant rate of $1 m m s^{- 1} .$ The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be _______ (round off to nearest integer)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$15 µ V$

$50 µ V$

$5 µ V$

$25 µ V$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The data given is $B = 0.4 T$
$R = 2 \times 10^{- 2} m$
It is given that $\frac{d R}{d t} = 10^{- 3} m s^{- 1}$
Area of the loop is $A = π R^{2}$
So, $\frac{d A}{d t} = 2 π R \frac{d R}{d t}$
The flux is, $ϕ = 2 π R \frac{d R}{d t} \times B$
So, the magnitude of emf is
$ε = 2 \times \frac{22}{7} \times 2 \times 10^{- 2} \times 10^{- 3} \times 0.4$
$= 5.028 \times 10^{- 5} V = 50.28 μ V \approx 50 μ V$

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET